Question

Derivada $F(t)= \frac{ \sqrt{ e^{t} -1} }{ { \sqrt{ e^{t} } } +1}$

Answer 1

$f'(t)= \dfrac{ \sqrt{e^t-1} }{ \sqrt{e^t+1} }\\\\\\Regra\,\,do\,\,Quociente\\\\f'(t)= \dfrac{f'(t)\cdot{g(t)}-f(t)\cdot{g'(t)}}{[g(t)]^2} \\\\\\f(t)=(e^t-1)^{\frac{1}{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g(t)=(e^t+1)^{\frac{1}{2}}\\f'(t)= \dfrac{e^t}{2}(e^t-1)^{-\frac{1}{2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g'(t)= \dfrac{e^t}{2}(e^t+1)^{-\frac{1}{2} }$


$f'(t)= \dfrac{\dfrac{e^t}{2}\cdot{(e^t-1)^{-\frac{1}{2}}}-[(e^t-1)^{{\frac{1}{2}}}\cdot{ \dfrac{e^t}{2}(e^t+1)^{-\frac{1}{2}} }]}{[(e^t+1)^{\frac{1}{2}}]^2} \\\\\\f'(t)= \dfrac{ \dfrac{e^t}{2}\cdot{(e^t-1)^{-\frac{1}{2}}}- \dfrac{e^t}{2}\cdot{(e^t-1)^{\frac{1}{2}})\cdot{(e^t+1)^{-\frac{1}{2}}}} }{e^t+1}$


$f'(t)= \dfrac{e^t\cdot{ \sqrt{(e^t+1}) }}{2\cdot{(e^t+1)\cdot \sqrt{e^t-1} }} - \dfrac{e^t\cdot{ \sqrt{e^t-1} }}{2\cdot(e^t+1)\cdot \sqrt{e^t+1} }$


$f'(t)= \dfrac{e^t\cdot \sqrt{e^t+1}\cdot \sqrt{e^t+1}-e^t\cdot \sqrt{e^t-1}\cdot \sqrt{e^t-1} }{2\cdot(e^t+1)\cdot \sqrt{e^t-1}\cdot \sqrt{e^t+1} } \\\\\\f'(t)= \dfrac{e^t\cdot(e^t+1)-e^t\cdot(e^t-1)}{2\cdot(e^t+1)^{\frac{3}{2}}\cdot \sqrt{e^t-1} } \\\\\\f'(t)= \dfrac{e^t\cdot(\diagup\!\!\!\!\!\!{e^t}+1-\diagup\!\!\!\!\!\!{e^t+1})}{2\cdot(e^t+1)^{\frac{3}{2}}\cdot \sqrt{e^t-1} }\\\\\\f'(t)=\dfrac{\diagup\!\!\!\!\!{2}\,e^t}{\diagup\!\!\!\!\!{2}\cdot(e^t+1)^{\frac{3}{2}}\cdot \sqrt{e^t-1} }$


$\boxed{f'(t)= \dfrac{e^t}{2\cdot(e^t+1)^{\frac{3}{2}}\cdot(e^t-1)\cdot \sqrt{e^t-1} } }$